#$&* course Mth 158 7/10 330 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6). The function then decreases to its local min at (5, 0), after which it continues increasing. So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand why the decreasing of the graph from (-2, 6) to (0,0) isn’t included in the solution? The graph shows it decreasing at those points. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: * * The function intersects the x axis at (-1, 0) and (1, 0), and the y axis at (0, 2). The function decreases from (-3,3) to (-1,0) so it is decreasing on the interval (-3, -1). The function decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1). The function increases from (-1,0) to (0, 2) so it is increasing on the interval (-1, 0). The function ioncreases from (1,0) to (3, 3) so it is increasing on the interval (1, 3). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is 0 <= y <= 3, written as the interval [0, 3]. The function is symmetric about the y axis, with f(-x) = f(x) for every x (e.g., f(-3) = f(3) = 3; f(-1) = f(1) = 0; etc.). So the function is even. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Would my range and domain count? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: * * The function intersects the x axis at (-2.25, 0) and (3, 0), and the y axis at (0, 1). The function decreases from (2,2) to (3,0) so it is decreasing on the interval (2,3). The function increases from (-3,-2) to (-2, 1) so it is increasing on the interval (-3, -2). The function ioncreases from (0, 1) to (2, 2) so it is increasing on the interval (0, 2). The function value does not change between (-2, 1) and (0, 1) so the function is constant on (-2, 0). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is -2 <= y <= 2, written as the interval [-2, 2]. The function is not symmetric about the y axis, with f(-x) not equal to f(x) for many values of x; e.g., f(-3) is -2 and f(3) is 0. So the function is not even. x = 3 shows us that the function is not odd either, since for an odd function f(-x) = -f(x) and for x = 3 this is not the case. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to include my domain and range. Also, my book tells me that the x intercept is (-2.3, 0) instead of (-2.25, 0).
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Given Solution: * * Local maximum is (0,1) Local minimum are (-pi,-1) and (pi,-1) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 3.3.76 / 46 (was 3.2.36) (f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2 What is your expression for (f(x) - f(1) ) / (x - 1) and how did you get this expression? How did you use your result to get the ave rate of change from x = 1 to x = 2, and what is your value? What is the equation of the secant line from the x = 1 point to the x = 2 point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-2x^2) - (-1) which is -2x^2 + x + 1 Factors to (2x + 1) (-x +1) or (x-1) So (2x + 1)(x-1) / (x-1) Slope of the secant line would equal (2x +1) To find the average rate of change: (1-2)^2 - (2-2)^2 / 2-1 Equals 1. I know that you are supposed to plug the average rate of change into the point slope formula to find the equation of the secant line but I don’t know what my two points would be. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * f(x) - f(1) = (x-2x^2) - (-1) = -2 x^2 + x + 1. This factors into (2x + 1) ( -x + 1). Since -x + 1 = - ( x - 1) we obtain (f(x) - f(1) ) / ( x - 1) = (2x + 1) ( -x + 1) / (x - 1) = (2x + 1) * -1 = - (2x + 1). A secant line runs from one point of the graph to another. The secant line here runs from the graph point (1, f(1) ) to the graph ponit ( x , f(x) ), and the expression we have just obtained is the slope of the secant line. For x = 2 the expression -(2x + 1) gives us ( f(2) - f(1) ) / ( 2 - 1), which is the slope of the line from (1, f(1) ) to (2, f(2)) . -(2 * 2 + 1) = -5, which is the desired slope. The secant line contains the point (1, -1) and has slope 5. So the equation of the secant line is (y - (-1) ) = -5 * (x - 1), which we solve to obtain y = -5 x + 4. ** STUDENT QUESTION 100724 I'm OK up to the point where ( f(x) - f(1) ) / (x - 1) = - ( 2 x + 1 ). Beyond that, I don't know where to go and can't follow the given solution. INSTRUCTOR RESPONSE You appear to understand the expression (f(x) - f(1) ) / (x - 1), and you appear to know how to simplify the resulting numerator, factor, and simplify the resulting expression. That's a very good start. Optional series of exercises: Here's a numerical exercise, using the same function. What is numerical value of the expression (f(3) - f(1) ) / (3 - 1)? &&&& What are the x = 1 and x = 3 points of a graph of f(x) vs. x? &&&& Sketch a set of coordinate axes, and plot these two points. Connect the two points with a straight line segment. Describe your sketch. &&&& What is the rise from the first point to the second? &&&& What is the run from the first point to the second? &&&& What therefore is the slope of your line segment? &&&& Between x = 1 and x = 3, do you think the graph of the actual function f(x) is a straight line? Why or why not? &&&& Your line segment went from one point of the f(x) graph to another. A line segment from one point on a curve to another is called a secant. So you have just calculated the slope of a secant line. Restate this in your own words: &&&& Your secant is a straight line segment, and you have just found its slope. What are the coordinates of the x = 1 point? &&&& You have found the slope of the secant, and you have just given the coordinates of a point on that secant. How do you find the equation of a straight line when you know its slope and the coordinates of a point on the line? &&&& What therefore is the equation of the secant line? &&&& Now repeat the same reasoning to do the following: • Find the x = 1 and x = 2 points of the graph. The x = 1 point will be the same as before. Then find the slope of the secant line between the x = 1 and x = 2 points. Finally find the equation of the secant line and simplify it to the form y = m x + b. Having gone through the process for these points, you should obtain the equation y = -5 x + 4. Now, since the present function is f(x) = x - 2 x^2, the point (x, f(x)) has coordinates ( x, x - 2 x^2). What were the coordinates of your x = 1 point again? &&&& Between your x = 1 point and the general point (x, x - 2 x^2): What are the two x coordinates of your points? (The coordinates of the second point both contain x's). &&&& What therefore is the expression for the run between these points? (there will be x's in this expression also) &&&& What are the two y coordinates of your points? (yes, there will be x's in this expression .. ) &&&& What therefore is the expression for the slope between these points? (and it will be no surprise to find x's in this expression) &&&& A line segment from your x = 1 point to the point (x, x - 2 x^2) is a secant of the graph of f(x) = x - 2 x^2. You have just found the slope of the secant. Explain this in your own words: &&&& Your expression for the secant included the variable x. If you replace x by 2 and evaluate your expression for the secant, what do you get? &&&& What were the coordinates of your x = 1 point again? &&&& What was the expression you got for the slope of the secant line when you replaced x by 2? &&&& What therefore is the equation of the secant line through your x = 1 point, when x has been replaced by 2? &&&& What does all this tell you about the given solution to the problem? &&&& If you had trouble understanding the second half of the given solution, you should consider submitting the optional exercise. If you do, insert each answer in the line that contains the &&&&, but before the &&&& mark, so that your insertion ends with &&&&. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like I am missing something. I don’t understand what you mean by finding the numerical value of the expression (f(3) - f(1) ) / (3 - 1)? ------------------------------------------------ Self-critique Rating:
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Given Solution: * * h(x) = 3x^3 +5 = -3x^3 + 5 h(x) is not equal to h(-x), which means that the function is not even. h(x) is not equal to -h(-x), since 3 x^3 + 5 is not equal to - ( -3x^3 + 5) = 3 x^3 - 5. So the function is not odd either. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After looking at your solution, I am still confused as to what you did. Why did the problem go from being 3 times x^3 to 3 - x^3? I do know how to tell by looking at a graph if the function is odd, even, or neither but how do you tell from just having an equation? ------------------------------------------------ Self-critique Rating: 0